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CHE 294 S01 Homework 7 Solution 21.June.
restart:
Problem 1
part a
Shift Axes Solution
Shifting the xyz axes to x'y'z' as shown below allows us to determine the intercepts immediately.
x
y
z
x’
y’
z’
The intercepts are X=1, Y=1/2, Z=1. Invert and clear fractions.
this gives the (121) plane
part b
Matrix Algebra Solution
Solve simultaneous equations using linear matrix algebra.
pA := Matrix([[0,1,1],[-1,2,0],[1,1,0]]); # matrix of points
I1 := Vector([1,1,1]); # unit vector
pA :=
I1 :=
Solution is found by taking inverse of point matrix multiplied by unit vector.
with(LinearAlgebra): MI := MatrixInverse(pA).I1;
MI :=
clearing fractions gives the (121) plane
Simultaneous Equations Solution
Solve simultaneous equations directly.
EqnMI1 := h0 + k1 + l1 = 1: EqnMI2 := h(-1) + k2 + l0 = 1:
EqnMI3 := h1 + k1 + l*0 = 1: solve({EqnMI1,EqnMI2,EqnMI3},{h,k,l});
{ h = , , }
l =
k =
clearing fractions gives the (121) plane
part c
Cross Product Solution
Find two vectors in the plane (along its edges) and take the cross product.
VectorA := <pA[2,1]-pA[1,1],pA[2,2]-pA[1,2],pA[2,3]-pA[1,3]>;
VectorB := <pA[3,1]-pA[1,1],pA[3,2]-pA[1,2],pA[3,3]-pA[1,3]>;
VectorA :=
VectorB :=
CrossProduct(VectorA,VectorB);
CrossProduct(VectorB,VectorA);
The cross product of the two vectors gives the Miller index of a vector perpendicular to the plane.
The plane and its negative are the same plane.
the plane is the (121)
FCC (111) Plane
The (111) plane has one atom in a parallelogram unit cell with sides of length 2r and 60 degree
minor angles. The area of the unit cell is a b sin( θ )where a and b are the lengths of the sides
adjoining the minor angle. This gives:
Warning, premature end of input
AFCC111unitcell := (2r)(2r)sin(60*Pi/180);
AFCC111unitcell := 2 r
2 3
SurDenFCC111 := 1/AFCC111unitcell:
FCC (100) Plane
The (100) plane has two atoms in a square. The lattice parameter is related to radius by
2 ao = 4 r since the atoms touch on the face diagonal. This gives:
Warning, premature end of input
AFCC100unitcell := ao^2: ao := 4*r/sqrt(2): AFCC100unitcell;
8 r
2
SurDenFCC100 := 2/AFCC100unitcell:
Solution for Pt
Substitute the radius of Pt (in cm) and solve.
SPt1111 := evalf3;
SPt1111 :=.151 10
16
SPt100 := evalf3;
SPt100 :=.131 10
16
surface density of Pt(111) is about 1.5 x 10E15 atoms/cm
surface density of Pt(100) is about 1.3 x 10E15 atoms/cm
Problems 3 and 4 are due on Assignment 8.
